With C an arbitrary constant, find a solution to the differential equation given: (csc x)y' + csc y = 0?

a. y = Cx^-4





b. y = Cx^-1/4





c. y = Cx^1/4





d. y = Cx^4





e. sin y + cos x = C





f. cos y + sin x = C





g. cos y - sin x = C





h. sin y - cos x = C





or is it none of these ?

With C an arbitrary constant, find a solution to the differential equation given: (csc x)y' + csc y = 0?
I would say none of the above. You can check my work.





You want to integrate. To do this we need to notice first that y' = dy/dx so





csc(x)(dy/dx) + csc(y) = 0





Multiply both sides by dx





csc(x)dy + csc(y)dx = 0





Divide both sides by csc(x)csc(y)





dy/csc(y) + dx/csc(x) = 0 eqn1





We know csc(z) = 1/sin(z) so eqn1 turns into





sin(y)dy + sin(x)dx = 0





Move x term to other side





sin(y)dy = -sin(x)dx





Take the integral on both sides





Ssin(y)dy = -Ssin(x)dx


-cos(y) + C1 = cos(x) + C2





Putting all constants together


-cos(y) = cos(x) +C3


or


cos(x) + cos(y) = -C3 = C


I differentiate between the constants because some people get confused if I don't.